graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y ) and ( y ), finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. identify any symmetries. choose the correct answer below. a. the function is an odd function that is symmetric about the y - axis. b. the function is an even function that is symmetric about the y - axis. c. the function is an even function that is symmetric about the origin. d. the function is an odd function that is symmetric about the origin. e. the function is neither even nor odd. find the derivative ( y ). ( y=square )

Question

Accepted Answer

# Explanation:
## Step1: Determine symmetry
For a function $y = f(x)=\frac{2x}{x^{2}-16}$, check $f(-x)$.
$$
\begin{align*}
f(-x)&=\frac{2(-x)}{(-x)^{2}-16}\
&=\frac{- 2x}{x^{2}-16}\
&=-f(x)
\end{align*}
$$
Since $f(-x)=-f(x)$, the function is odd. Odd - functions are symmetric about the origin.

## Step2: Find the derivative $y'$ using the quotient rule
If $y = \frac{u}{v}$ where $u = 2x$ and $v=x^{2}-16$, the quotient rule states that $y'=\frac{u'v - uv'}{v^{2}}$
- $u'=\frac{d}{dx}(2x)=2$
- $v'=\frac{d}{dx}(x^{2}-16)=2x$

$$
\begin{align*}
y'&=\frac{2(x^{2}-16)-2x(2x)}{(x^{2}-16)^{2}}\
&=\frac{2x^{2}-32 - 4x^{2}}{(x^{2}-16)^{2}}\
&=\frac{-2x^{2}-32}{(x^{2}-16)^{2}}\
&=\frac{-2(x^{2}+16)}{(x^{2}-16)^{2}}
\end{align*}
$$

# Answer:
The function is an odd function that is symmetric about the origin (Option D). The derivative $y'=\frac{-2(x^{2}+16)}{(x^{2}-16)^{2}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Determine symmetry

Step2: Find the derivative \(y'\) using the quotient rule

Answer: